This is just the following result.
Lemma 33.23.1. Let k be an algebraically closed field. Let X be a proper k-scheme. Let \mathcal{L} be an invertible \mathcal{O}_ X-module. Let V \subset H^0(X, \mathcal{L}) be a k-subvector space. If
for every pair of distinct closed points x, y \in X there is a section s \in V which vanishes at x but not at y, and
for every closed point x \in X and nonzero tangent vector \theta \in T_{X/k, x} there exists a section s \in V which vanishes at x but whose pullback by \theta is nonzero,
then \mathcal{L} is very ample and the canonical morphism \varphi _{\mathcal{L}, V} : X \to \mathbf{P}(V) is a closed immersion.
Proof.
Condition (1) implies in particular that the elements of V generate \mathcal{L} over X. Hence we get a canonical morphism
\varphi = \varphi _{\mathcal{L}, V} : X \longrightarrow \mathbf{P}(V)
by Constructions, Example 27.21.2. The morphism \varphi is proper by Morphisms, Lemma 29.41.7. By (1) the map \varphi is injective on closed points (computation omitted). In particular, the fibre over any closed point of \mathbf{P}(V) is a singleton (small detail omitted). Thus we see that \varphi is finite, for example use Cohomology of Schemes, Lemma 30.21.2. To finish the proof it suffices to show that the map
\varphi ^\sharp : \mathcal{O}_{\mathbf{P}(V)} \longrightarrow \varphi _*\mathcal{O}_ X
is surjective. This we may check on stalks at closed points. Let x \in X be a closed point with image the closed point p = \varphi (x) \in \mathbf{P}(V). Since \varphi ^{-1}(\{ p\} ) = \{ x\} by (1) and since \varphi is proper (hence closed), we see that \varphi ^{-1}(U) runs through a fundamental system of open neighbourhoods of x as U runs through a fundamental system of open neighbourhoods of p. We conclude that on stalks at p we obtain the map
\varphi ^\sharp _ x : \mathcal{O}_{\mathbf{P}(V), p} \longrightarrow \mathcal{O}_{X, x}
In particular, \mathcal{O}_{X, x} is a finite \mathcal{O}_{\mathbf{P}(V), p}-module. Moreover, the residue fields of x and p are equal to k (as k is algebraically closed – use the Hilbert Nullstellensatz). Finally, condition (2) implies that the map
T_{X/k, x} \longrightarrow T_{\mathbf{P}(V)/k, p}
is injective since any nonzero \theta in the kernel of this map couldn't possibly satisfy the conclusion of (2). In terms of the map of local rings above this means that
\mathfrak m_ p/\mathfrak m_ p^2 \longrightarrow \mathfrak m_ x/\mathfrak m_ x^2
is surjective, see Lemma 33.16.5. Now the proof is finished by applying Algebra, Lemma 10.20.3.
\square
Lemma 33.23.2. Let k be an algebraically closed field. Let X be a proper k-scheme. Let \mathcal{L} be an invertible \mathcal{O}_ X-module. Suppose that for every closed subscheme Z \subset X of dimension 0 and degree 2 over k the map
H^0(X, \mathcal{L}) \longrightarrow H^0(Z, \mathcal{L}|_ Z)
is surjective. Then \mathcal{L} is very ample on X over k.
Proof.
This is a reformulation of Lemma 33.23.1. Namely, given distinct closed points x, y \in X taking Z = x \cup y (viewed as closed subscheme) we get condition (1) of the lemma. And given a nonzero tangent vector \theta \in T_{X/k, x} the morphism \theta : \mathop{\mathrm{Spec}}(k[\epsilon ]) \to X is a closed immersion. Setting Z = \mathop{\mathrm{Im}}(\theta ) we obtain condition (2) of the lemma.
\square
Comments (2)
Comment #7149 by Manolis C. Tsakiris on
Comment #7299 by Johan on